What will be the acceleration due to gravity at a height $h$ above the Earth's surface,where $R$ is the radius of the Earth and $g$ is the acceleration due to gravity on the surface of the Earth?

The depth $d$ at which the acceleration due to gravity becomes $\frac{g}{n}$ is (where $R$ is the radius of the Earth,$g$ is the acceleration due to gravity at the surface,and $n$ is an integer).

Consider Earth to be a sphere of radius $R_e$ rotating about its own axis with angular speed $\omega$. If $g_{E}$ and $g_{P}$ are the accelerations due to gravity at the equator and the poles respectively,then $(g_{P}-g_{E})$ is given by $\left[\cos (0^{\circ})=\sin (\frac{\pi}{2})=1, \sin (0^{\circ})=\cos (\frac{\pi}{2})=0\right]$

What is the depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface of the Earth? (radius of Earth $= R$)

The mass of the moon is $(1/8)$ of the earth,but the gravitational pull is $(1/6)$ of the earth. This is due to the fact that:

$A$ body weighs $700\,gm\,wt.$ on the surface of the earth. How much will it weigh on the surface of a planet whose mass is $\frac{1}{7}$ and radius is half of that of the earth?